TITLE: The set of ultrafilters on an infinite set is uncountable
QUESTION [26 upvotes]: After recently learning about filters and ultrafilters, we looked into further problems and properties. I am having trouble with this one:
If $X$ is an infinite set, then the set of all ultrafilters on $X$ is uncountable.

REPLY [2 votes]: Here are some arguments which are not fundamentally different from the ones in the other answers, but are possibly made more enlightening by being cast in the language of topology.
If $X$ is uncountable, there are uncountably many principal ultrafilters on $X$, so we may assume $X$ is countable.  In particular, we can identify $X$ with $\mathbb{Q}$.  Now in any topological space $S$, if $D\subseteq S$ and $x\in\overline{S}$, there is an ultrafilter on $D$ which converges to $s$ (proof: take the filter of neighborhoods of $x$ intersected with $D$ and extend it to an ultrafilter).  In particular, considering $D=\mathbb{Q}$ and $S=\mathbb{R}$, for each $x\in\mathbb{R}$ there is an ultrafilter on $\mathbb{Q}$ converging to $x$.  Since $\mathbb{R}$ is Hausdorff, a filter cannot have more than one limit, and so these ultrafilters corresponding to each point of $\mathbb{R}$ must be distinct.  This gives uncountably many distinct ultrafilters on $\mathbb{Q}$.
More generally, the same reasoning shows that if $S$ is a Hausdorff space with a dense subset $D$, then there are at least $|S|$ different ultrafilters on $D$.  Given any set $X$, we can take $S=\{0,1\}^X$, and $D$ to be the set of points that are $0$ on all but finitely many coordinates.  If $X$ is infinite then $|D|=|X|$, and we conclude that there are at least $|S|=2^{|X|}$ ultrafilters on $X$.
We can use a similar but trickier construction to prove the sharp lower bound of $2^{2^{|X|}}$ ultrafilters.  Let $S=\{0,1\}^{\mathcal{P}(X)}$.  For each $x\in X$, we have a map $u_x:\mathcal{P}(X)\to \{0,1\}$ given by $u_x(A)=1$ iff $x\in A$.  Now let $D$ be the set of functions $\mathcal{P}(X)\to\{0,1\}$ which can be written as a finite pointwise Boolean combination of functions of the form $u_x$.  Clearly $|D|=|X|$ if $X$ is infinite, and I claim $D$ is dense in $S$.
To prove this, let $A_1,\dots,A_n\in\mathcal{P}(X)$.  We can find a finite subset $X_0\subset X$ such that $A_1\cap X_0,\dots, A_n\cap X_0$ are all distinct.  The functions $u_x$ for $x\in X_0$ generate the Boolean algebra of all functions $\mathcal{P}(X_0)\to \{0,1\}$ (the characteristic function of any singleton $\{B\}\subseteq\mathcal{P}(A_0)$ is just the join of all the $u_x$ for $x\in B$ and the negations of the $u_x$ for $x\not\in B$).  In particular, we can find a Boolean combination of the $u_x$ for $x\in X_0$ which takes arbitrary specified values on the sets $A_1\cap X_0,\dots,A_n\cap X_0$ and thus will take the same values on $A_1,\dots,A_n$.  This exactly proves that $D$ is dense in $S=\{0,1\}^{\mathcal{P}(X)}$ in the product topology.  Since $S$ is Hausdorff and has $2^{2^{|X|}}$ points, this shows there are $2^{2^{|X|}}$ ultrafilters on $D$ and hence on $X$.  (For more discussion of this construction, see Is $2^{\frak{c}}$ separable?.)
The arguments above are all in the spirit of Brian Scott's answer, just in the language of topology.  Here is a topological argument which is instead along the lines of Carl Mummert and Alex Ravsky's answers.  Note that the set of nonprincipal ultrafilters on a set $X$ can naturally be identified with the Stone space of the the Boolean algebra $\mathcal{P}(X)/\mathrm{fin}$, where $\mathrm{fin}$ is the ideal of finite subsets of $X$.  This Stone space is compact Hausdorff, nonempty if $X$ is infinite, and has no isolated points since $\mathcal{P}(X)/\mathrm{fin}$ is atomless (given an infinite subset of $X$, we can partition it into two infinite subsets).  But a nonempty compact Hausdorff space with no isolated points is uncountable, so there are uncountably many nonprincipal ultrafilters if $X$ is infinite.